LazyPromise implementation. Doesn't invoke the executor function until `.then()` is called.

example.js

'use strict';

const Lazy = require('./lazy-promise');

class TimerPromise extends Lazy(Promise) {
  constructor() {
    const start = new Date();
    super((resolve, reject) => resolve(new Date() - start));
  }
}


var timer = new TimerPromise();
console.log('created lazy timer promise');

setTimeout(() => timer.then((ms) => {
  console.log('%d ms elapsed before executor was called', ms)
}), 2000);

lazy-promise.js

'use strict';

const $executor = Symbol('Lazy#executor');
const $promise = Symbol('Lazy#promise');

const inherits = require('util').inherits;

function Lazy (Promise) {
  function LazyPromise(fn) {
    if (this[$promise]) {
      throw new Error('LazyPromise is already being resolved');
    }
    this[$executor] = fn;
    this[$promise] = null;

    // we intentionally do not call `Promise.call(this, fn)` here,
    // hence ES6 Classes cannot be used here since `super()` is required there
  }

  inherits(LazyPromise, Promise);

  LazyPromise.prototype.then = function then (resolve, reject) {
    var promise = this[$promise];
    if (!promise) promise = this[$promise] = new Promise(this[$executor]);
    return promise.then(resolve, reject);
  };

  return LazyPromise;
};

module.exports = Lazy;